Tamil Nadu Board of Secondary EducationHSC Commerce Class 12th

# Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem. Destination D1 D2 D3 D4 Supply O1 2 3 11 7 6 Origin O2 1 0 6 1 1 O3 5 8 15 - Business Mathematics and Statistics

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Sum

Explain Vogel’s approximation method by obtaining initial basic feasible solution of the following transportation problem.

 Destination D1 D2 D3 D4 Supply O1 2 3 11 7 6 Origin O2 1 0 6 1 1 O3 5 8 15 9 10 Demand 7 5 3 2

#### Solution

Total supply (ai) = 6 + 1 + 10 = 17

Total demand (bj) = 7 + 5 + 3 + 2 = 17

sum"a"_"i" = sum"b"_"j".

So given transportation problem is a balanced problem and hence we can find a basic feasible solution by VAM.

First allocation:

 D1 D2 D3 D4 (ai) penalty O1 2 3 11 7 6 (1) O2 1 0 6 (1)1 1/0 (1) O3 5 8 15 9 10 (3) (bj) 7 5 3 2/1 penalty (1) (3) (5) (6)

The largest penalty is 6.

So allot min (2, 1) to the cell (O2, D4) which has the least cost in column D4.

Second allocation:

 D1 D2 D3 D4 (ai) penalty O1 2 (5)3 11 7 6/1 (1) O3 5 8 15 9 10 (3) (bj) 7 5/0 3 1 penalty (3) (5) (4) (2)

The largest penalty is 5.

So allot min (5, 6) to the cell (O1, D2) which has the least cost in column D2.

Third allocation:

 D1 D3 D4 (ai) penalty O1 (1)2 11 7 1/0 (5) O3 5 15 9 10 (4) (bj) 7/6 3 1 penalty (3) (4) (2)

The largest penalty is 5.

So allot min (7, 1) to the cell (O1, D1) which has the least cost in row O1.

Fourth allocation:

 D1 D3 D4 (ai) penalty O3 (6)5 15 9 10/4 (4) (bj) 6/0 3 1 penalty – – –

Fifth allocation:

 D3 D4 (ai) penalty O3 (3)15 (1)9 4/3/0 (6) (bj) 3/0 1/0 penalty – –

The largest penalty is 6.

So allot min (1, 4) to the cell (O3, D4) which has the least cost in row O3.

The balance 3 units is allotted to the cell (O3, D3).

We get the final allocation as given below.

 D1 D2 D3 D4 Supply O1 (1)2 (5)3 11 7 6 O2 1 0 6 (1)1 1 O3 (6)5 8 (3)15 (1)9 10 Demand 7 5 3 2

Transportation schedule:

O1 → D1

O1 → D2

O2 → D4

O3 → D1

O3 → D3

O3 → D4

i.e x11 = 1

x12 = 5

x24 = 1

x31 = 6

x33 = 3

x34 = 1

Total cost is given by = (1 × 2) + (5 × 3) + (1 × 1) + (6 × 5) + (3 × 15) + (1 × 9)

= 2 + 15 + 1 + 30 + 45 + 9

= 102

Thus the optimal (minimal) cost of the given transportation problem by Vogel’s approximation method is ₹ 102.

Concept: Transportation Problem
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